# Inductor and Capacitor Questions

Discussion in 'Electronic Basics' started by JJD, Dec 25, 2003.

1. ### JJDGuest

Hello all, this is my first post.

C question:
I understand that the equation i = C*dv/dt tells me that if there is
an instantaneous change in voltage across a cap, that would require an
infinite amount of current through it.
If I charge a cap, remove it and "hold it", and put a dieletric
material, k, between the plates, then there is an instantaneous change
in voltage (decrease) right? q and C increase and V decreases, but i =
0. I know that any energy lost is applied to the physical motion of
the k material. How does all of this make sense?
i = C*dv/dt, q = C*V, C = k*Cair(the capacitance in air)

L question:
If I have a simple L, Vs and switch series circuit, when I open the
switch there should be an infinite V across the open switch, there is
arching for a split second, but how much V is there across the switch
in real life?

2. ### John LarkinGuest

Only if you 'put' instantaneously.
Sounds right; i = 0 by definition, since nothing is connected to the
capacitor leads, which is where you'd measure i. What part doesn't
make sense?

Depends on lots of things: breakdown properties of the gap created as
the switch opens, circuit distributed capacitance, rate of switch
opening, amount of stored energy, stuff like that. Hundreds of volts
peak would not be unusual with Radio Shack parts, multi kilovolts
could be achieved with some effort. If distributed capacitance is high
enough, it might not arc at all... it would oscillate, with amplitude
tapering off as the energy is dissipated in various ways.

Get some parts and try it!

John

3. ### John PopelishGuest

The capacitance changes but the charge is unchanged (no current in or
out), so the voltage changes.
As much as it takes to allow the current through the inductor to decay
in a continuous manner. Depending on how fast the contacts separate,
the peak voltage may be in the hundreds of volts or thousands. If the
contacts separate fast enough, the arc will extinguish before the
current reaches zero, and the capacitance between the windings of the
inductor will carry the inductive current, ringing with the inductance
in a decaying sinusoid as energy dissipates in the losses in the
inductance and capacitance and as energy radiating away from the

There's a fun way of demonstrating this. Get an air-dielectric
variable capacitor as was used as the tuning capacitor in old (vacuum
charge it up to a couple of hundred volts or so (below the voltage
where it arcs), disconnect the charging source, and turn the shaft to
lower the capacitance. The charge remains the same, but the
capacitance decreases, so the voltage increases until it arcs over.

5. ### MacGuest

I = C dv/dt applies to constant C caps.

When you insert dielectric between the plates of a parallel plate
capacitor, you alter C. You cannot alter q by doing this, however.

If inserting the dielectric leads to a capacitor with more potential
energy in it, then that means you will have to apply mechanical force to
get the dielectric in place. And vice verse.
The voltage will continue to rise until current flows or the backwards
EMF is enough to shut down the current. In practice, as you seem to know,
something always gives, in your case, the switch. So, the voltage is
clamped by the breakdown voltage of the switch, which is probably not very
well characterized and in any event, depends on the atmosphere around the
switch.

In cases where you don't want the switch to arc, you may put a
free-wheeling diode in parallel with the inductor. The parallel diode only
works when the current direction is always the same through the inductor.

Diodes don't turn on instantly, so you will still get some voltage spiking
when the switch is opened.

Mac

6. ### John WoodgateGuest

Q doesn't increase. You can tell from I = dQ/dt.
It depends on how fast the contacts separate. During the arc, there is
almost NO voltage across it. It could even be less that zero, because
arcs have negative incremental resistance characteristics.

7. ### Max_VeneziaGuest

when you calculate something keep in mind that "zero" and "infinite"
do not exist in the real phisical world. when time goes to zero the
modelling change from concentrate parameters to distributed parameters
and others parameters give a secondary effect.

8. ### JJDGuest

Thank you all for the help. I am an EE student-probably you can
tell-just starting my circuits series. I have a better understanding
now.

9. ### Bert HickmanGuest

John Woodgate wrote:
Not quite... the negative arc characteristic only means that dV/dI is
negative: increasing the current flow tends to decrease the voltage drop
across the arc. However, the overall resistance of the arc will still
always be positive.

-- Bert --
--

10. ### John WoodgateGuest

I read in sci.electronics.design that Bert Hickman <[email protected]_
You don't even read, or can't understand, what I wrote. What do you
think 'incremental' means?

11. ### John LarkinGuest

Incremental negative resistance cannot produce absolute polarity
reversal; that violates conservation of energy. The Free Energy folk
are great fans of incremental negative slopes.

Oscillatory ringing could (generally does) result in polarity
reversal, but that's a whold nother issue.

John

12. ### John WoodgateGuest

I read in sci.electronics.design that John Larkin <[email protected]
4ax.com>) about 'Inductor and Capacitor Questions', on Sat, 27 Dec 2003:
Indeed. When I wrote 'less then zero' I was referring to the AC voltage.
In a DC circuit, of course, the arc voltage is always greater than zero,
but often not by much.
Like the Gadarene Swine. Or lemmings.(;-)
Indeed.

13. ### John LarkinGuest

What is a less-than-zero AC voltage? My Fluke doesn't have a negative
AC range.

John

14. ### John WoodgateGuest

I read in sci.electronics.design that John Larkin <[email protected]